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A whistle of frequency 1000 Hz is sounded on a car travelling towards a cliff with velocity of 18 m s–1 normal to the cliff. If c = 330 m s–1, then the apparent frequency of the echo as heard by the car driver is nearly
- a)1115 Hz
- b)115 Hz
- c)67 Hz
- d)47.2 Hz
Correct answer is option 'A'. Can you explain this answer?
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A whistle of frequency 1000 Hz is sounded on a car travelling towards ...
(a) By the concept of accoustic, the observer and source are moving towards each other, each with a velocity of 18 m s–1.
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A whistle of frequency 1000 Hz is sounded on a car travelling towards ...
-1
. The speed of sound in air is 340 m s-1. What is the frequency of the sound heard by an observer standing near the cliff?
We can use the Doppler effect formula to solve this problem:
f' = f(v + u) / (v + vs)
where:
f = frequency of the whistle (given as 1000 Hz)
v = speed of sound in air (given as 340 m s-1)
u = velocity of the car towards the cliff (given as 18 m s-1)
vs = velocity of the sound wave relative to the observer (what we need to find)
f' = frequency of the sound heard by the observer (what we need to find)
Plugging in the given values, we get:
f' = 1000 (340 + 18) / (340 + vs)
f' = 1020 / (340 + vs)
To find vs, we need to use the fact that the frequency of the sound wave is constant, so:
f = f' (v + vs) / (v + u)
Plugging in the given values and solving for vs, we get:
vs = (f - f'u) / (f' - f)
vs = (1000 - 1020(18)) / (1020 - 1000)
vs = 170 m s-1
Therefore, the frequency of the sound heard by the observer near the cliff is:
f' = 1020 / (340 + 170)
f' = 2/3 * 1000 Hz
f' = 666.67 Hz (rounded to two decimal places)
. The speed of sound in air is 340 m s-1. What is the frequency of the sound heard by an observer standing near the cliff?
We can use the Doppler effect formula to solve this problem:
f' = f(v + u) / (v + vs)
where:
f = frequency of the whistle (given as 1000 Hz)
v = speed of sound in air (given as 340 m s-1)
u = velocity of the car towards the cliff (given as 18 m s-1)
vs = velocity of the sound wave relative to the observer (what we need to find)
f' = frequency of the sound heard by the observer (what we need to find)
Plugging in the given values, we get:
f' = 1000 (340 + 18) / (340 + vs)
f' = 1020 / (340 + vs)
To find vs, we need to use the fact that the frequency of the sound wave is constant, so:
f = f' (v + vs) / (v + u)
Plugging in the given values and solving for vs, we get:
vs = (f - f'u) / (f' - f)
vs = (1000 - 1020(18)) / (1020 - 1000)
vs = 170 m s-1
Therefore, the frequency of the sound heard by the observer near the cliff is:
f' = 1020 / (340 + 170)
f' = 2/3 * 1000 Hz
f' = 666.67 Hz (rounded to two decimal places)
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A whistle of frequency 1000 Hz is sounded on a car travelling towards a cliff with velocity of 18 m s–1 normal to the cliff. If c = 330 m s–1, then the apparent frequency of the echo as heard by the car driver is nearlya)1115 Hzb)115 Hzc)67 Hzd)47.2 HzCorrect answer is option 'A'. Can you explain this answer?
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